The locus of the point given by the equations | vedclass

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(B) Given equations are $x = \frac{2at}{1 + t^2}$ and $y = \frac{a(1 - t^2)}{1 + t^2}$.Squaring both equations,we get $x^2 = \frac{4a^2t^2}{(1 + t^2)^

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Circle

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(B) Given equations are $x = \frac{2at}{1 + t^2}$ and $y = \frac{a(1 - t^2)}{1 + t^2}$.Squaring both equations,we get $x^2 = \frac{4a^2t^2}{(1 + t^2)^2}$ and $y^2 = \frac{a^2(1 - t^2)^2}{(1 + t^2)^2}$.Adding $x^2$ and $y^2$,we have $x^2 + y^2 = \frac{4a^2t^2 + a^2(1 - 2t^2 + t^4)}{(1 + t^2)^2}$.$x^2 + y^2 = \frac{a^2(4t^2 + 1 - 2t^2 + t^4)}{(1 + t^2)^2} = \frac{a^2(t^4 + 2t^2 + 1)}{(1 + t^2)^2}$.Since $t^4 + 2t^2 + 1 = (1 + t^2)^2$,we get $x^2 + y^2 = \frac{a^2(1 + t^2)^2}{(1 + t^2)^2} = a^2$.This represents a circle with radius $a$ centered at the origin.

The locus of the point given by the equations $x = \frac{2at}{1 + t^2}$ and $y = \frac{a(1 - t^2)}{1 + t^2}$ for $-1 \le t \le 1$ is a

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